Thread: Concavity and second derivatives

1. Concavity and second derivatives

How can I find the point(s) of inflection of the graph of the function:

$f(t)= (1-t)(t-4)(t^2-4)$

How can I approach this problem and how would I go about solving down for a final answer ?

2. distribute it out

find f'(t), then find f''(t)

once you have the second derivative, what does this tell you about finding inflection points?

Show us your work so far so we can help you find your mistakes!

3. Hi,

Thanks for your reply.

I am not sure what this would tell me -- this is why I'm posting here, I'm trying to learn from what you guys can tell me.

Thanks again for your reply though!!

4. Inflection points occur where the value of the second derivative changes sign (positive to negative or negative to positive). So find an equation for the second derivative; find where it is equal to zero, and make sure that it changes sign at that location.

It looks like f'' will be a polynomial so it should be easy to find where it equals zero

5. Hi guys,

I'm still having trouble with this question, I cannot seem to differentiate it and solve down and set to zero, etc.

I would very much appreciate some further help with this question.

EDIT:

I Found:

$t'=-4x^3+15x^2-20$

$t''=-12x^2+30x$

thus:

At zero x = 0, 5/2

What do I do from here? Also can someone verify this for me?

6. Originally Posted by jimmyp
Hi guys,

I'm still having trouble with this question, I cannot seem to differentiate it and solve down and set to zero, etc.

I would very much appreciate some further help with this question.

EDIT:

I Found:

$t'=-4x^3+15x^2-20$

$t''=-12x^2+30x$

thus:

At zero x = 0, 5/2

What do I do from here? Also can someone verify this for me?
$f'(t) = -4t^3+15t^2-20
$

$f''(t) = -12t^2+30t$

set $f''(t) = 0$ ...

$-6t(2t - 5) = 0$

$t = 0$ , $t = \frac{5}{2}$

inflection point at $t = 0$ because $f''(t)$ changes sign at $t = 0
$

inflection point at $t = \frac{5}{2}$ because $f''(t)$ changes sign at $t = \frac{5}{2}$